We know the familiar theorem for the change of variables involving random variables (see for example here, Theorem 4.1):

Suppose $T:(x,y) \mapsto (u,v)$ is a 1-1 mapping from some domain $D\subseteq \mathbb{R}^2$ to some range $R \subseteq \mathbb{R}^2$. Define the Jacobian $J$ as a function of $u$ and $v$ by $J(u,v) = \left| x_u y_v - x_vy_u\right|$. Assume the partials exist and are continuous. If $X,Y$ have joint probability density function $f_{X,Y}$, then the random variables $U,V$defined by $(U,V) =T(X,Y)$ are jointly continuous with joint probability density function $f_{U,V}$ given by $f_{U,V}(u,v) = f_{X,Y}(x(u,v), y(u,v))J(u,v)$ if $(u,v)\in R$ and $f_{U,V}(u,v) = 0$ otherwise.

Okay, now suppose $X,Y \sim \mathrm{Unif}[0,1]$ independent random variables, let $T:\mathbb{R}^2 \longrightarrow \mathbb{R}^2$, be given by $(x,y) \mapsto (x+y, x-y) =: (u,v)$. We have $D = [0,1]^2$, $R = [0,2]\times[-1,1]$ and $x(u,v) = \frac{u+v}{2}$ as well as $y(u,v) = \frac{u-v}{2}$. We obtain our Jacobian as $J(u,v) =\left| \frac{1}{2} \frac{-1}{2} - \frac{1}{2}\frac{1}{2}\right| = \left| \frac{-1}{4} - \frac{1}{4} \right| = \frac{1}{2}$. As clearly $f_{X,Y}(x,y) = f_X(x)f_Y(y) = (1)(1) = 1$ by independence, we get by our formula that $f_{U,V}(u,v) = \frac{1}{2}$ on $R$ and $0$ everywhere else. A simple integration gives us $\int_R f_{U,V}(u,v)\mathrm{d}u\mathrm{d}v = 2$ which is clearly a contradiction as p.d.f's integrate to $1$.

Can someone help me and tell me where I made a mistake? I feel like this is such an easy question, but I can't really see what I'm doing wrong... Any help will be much appreciated!


The joint pdf of $U$ and $V$ does not have support $R = [0,2]\times [-1,1]$ as you claim it does; the support is a square region where the square has vertices $(0,0)$, $(1,1)$, $(2,0)$ and $(1,-1)$. The area is $2$ and the pdf has value $\frac 12$ on the region.


$$f_{U,V}(u,v) = f_{X,Y}(\frac{u+v}{2},\frac{u-v}{2}).\frac{1}{2}$$ $$f_{X,Y}(x,y) = f_X{(x)}.f_Y{(y)}$$ $$ f_{U,V}(u,v) = \frac{1}{2}, 0\le u \le2, -1\le v\le1$$ $$ \left[\int_{-1}^{0}\int_{-u}^{2+u} + \int_{0}^{1}\int_{u}^{2-u}\right]\frac{1}{2}dvdu = 1; $$

It is indeed 1.

The mistake is in your limits, you took u to be from 0 to 2 and v from -1 to 1 and you got 2 two's and a pdf of one half resulting in 2.

  • $\begingroup$ What is W, Z? What are you integrating? My integral is surely 2, as f is constant 1/2 on a square of area 4. We get 4*1/2 = 2 not 1. $\endgroup$ – Marcel S Sep 4 '17 at 17:09
  • $\begingroup$ X and Y gets transformed, you may have to review deriving joint probabillity density methodology. $\endgroup$ – Satish Ramanathan Sep 4 '17 at 17:38
  • $\begingroup$ sorry, I changed the solution. I was interrupted in the middle while I was writing the solution. I apologize for being shamelessly confident on a wrong solution. $\endgroup$ – Satish Ramanathan Sep 4 '17 at 19:36
  • $\begingroup$ The calculations are incorrect starting from the support of $f_{U,V}$ which is not as stated, and the integral does not match the alleged support. $\endgroup$ – Dilip Sarwate Sep 4 '17 at 19:41
  • $\begingroup$ Now the calculations are correct and supports the support. Thanks for correcting $\endgroup$ – Satish Ramanathan Sep 4 '17 at 20:06

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